Chapter 24 Physical Pendulum 24.I Introduction.. 24.1.1 Simple Pendulum:Torque Approach..... 24.2 Physical Pendulum. 24.3 Worked Examples4 Example 24.1 Oscillating Rod..4 Example 24.3 Torsional Oscillator Example 24.4 Compound Physical Pendulum.9 Appendix 24A Higher-Order Corrections to the Period for Larger Amplitudes of a Simple Pendulum. 12
Chapter 24 Physical Pendulum 24.1 Introduction........................................................................................................... 1 24.1.1 Simple Pendulum: Torque Approach .......................................................... 1 24.2 Physical Pendulum................................................................................................ 2 24.3 Worked Examples................................................................................................. 4 Example 24.1 Oscillating Rod.................................................................................. 4 Example 24.3 Torsional Oscillator .......................................................................... 7 Example 24.4 Compound Physical Pendulum........................................................ 9 Appendix 24A Higher-Order Corrections to the Period for Larger Amplitudes of a Simple Pendulum ..................................................................................................... 12
Chapter 24 Physical Pendulum ...I had along with me....the Descriptions,with some Drawings of the principal Parts of the Pendulum-Clock which I had made,and as also of them of my then intended Timekeeper for the Longitude at Sea. John Harrison 24.1 Introduction We have already used Newton's Second Law or Conservation of Energy to analyze systems like the spring-object system that oscillate.We shall now use torque and the rotational equation of motion to study oscillating systems like pendulums and torsional springs 24.1.1 Simple Pendulum:Torque Approach Recall the simple pendulum from Chapter 23.3.1.The coordinate system and force diagram for the simple pendulum is shown in Figure 24.1. PR (t 0t)、 mg (a) (b) Figure 24.1 (a)Coordinate system and (b)torque diagram for simple pendulum The torque about the pivot point P is given by 元p=ipm×mg=1f×mg(cos8f-sine)=-Imgsin0k (24.1.1) The z-component of the torque about point P (p).=-mglsin0. (24.1.2) J.Harrison,A Description Concerning Such Mechanisms as will Afford a Nice,or True Mensuration of Time;...(London,1775),p.19. 24-1
24-1 Chapter 24 Physical Pendulum …. I had along with me….the Descriptions, with some Drawings of the principal Parts of the Pendulum-Clock which I had made, and as also of them of my then intended Timekeeper for the Longitude at Sea. 1 John Harrison 24.1 Introduction We have already used Newton’s Second Law or Conservation of Energy to analyze systems like the spring-object system that oscillate. We shall now use torque and the rotational equation of motion to study oscillating systems like pendulums and torsional springs. 24.1.1 Simple Pendulum: Torque Approach Recall the simple pendulum from Chapter 23.3.1.The coordinate system and force diagram for the simple pendulum is shown in Figure 24.1. (a) (b) Figure 24.1 (a) Coordinate system and (b) torque diagram for simple pendulum The torque about the pivot point P is given by τ P = rP, m × m g = l rˆ × mg(cosθ rˆ − sinθ ˆ θ) = −l mg sinθ kˆ (24.1.1) The z -component of the torque about point P (τ P )z = −mglsinθ . (24.1.2) 1 J. Harrison, A Description Concerning Such Mechanisms as will Afford a Nice, or True Mensuration of Time;…(London, 1775), p.19
When >0,(and the torque about P is directed in the negative k-direction (into the plane of Figure 24.1b)when 0 and the torque about P is directed in the positive k-direction(out of the plane of Figure 24.1b).The moment of inertia of a point mass about the pivot point P is I=m12.The rotational equation of motion is then ,l.=1a,=l, do (24.1.3) -mglsine ml2 de dr Thus we have do=_Esine, (24.1.4) agreeing with Eq.23.3.14.When the angle of oscillation is small,we may use the small angle approximation sin0三0， (24.1.5) and Eq.(24.1.4)reduces to the simple harmonic oscillator equation d' dt2 -86 1 (24.1.6) We have already studied the solutions to this equation in Chapter 23.3.A procedure for determining the period when the small angle approximation does not hold is given in Appendix 24A. 24.2 Physical Pendulum A physical pendulum consists of a rigid body that undergoes fixed axis rotation about a fixed point S (Figure 24.2). pivot S cm <cm Figure 24.2 Physical pendulum 24-2
24-2 When θ > 0 , (τ P )z 0 and the torque about P is directed in the positive kˆ -direction (out of the plane of Figure 24.1b). The moment of inertia of a point mass about the pivot point P is IP = ml 2 . The rotational equation of motion is then (τ P )z = IPα z ≡ IP d2 θ dt 2 −mglsinθ = ml 2 d2 θ dt 2 . (24.1.3) Thus we have 2 2 sin d g dt l θ = − θ , (24.1.4) agreeing with Eq. 23. 3.14. When the angle of oscillation is small, we may use the small angle approximation sinθ ≅ θ , (24.1.5) and Eq. (24.1.4) reduces to the simple harmonic oscillator equation 2 2 d g dt l θ ≅ − θ . (24.1.6) We have already studied the solutions to this equation in Chapter 23.3. A procedure for determining the period when the small angle approximation does not hold is given in Appendix 24A. 24.2 Physical Pendulum A physical pendulum consists of a rigid body that undergoes fixed axis rotation about a fixed point S (Figure 24.2). Figure 24.2 Physical pendulum
The gravitational force acts at the center of mass of the physical pendulum.Denote the distance of the center of mass to the pivot point S by lem.The torque analysis is nearly identical to the simple pendulum.The torque about the pivot point S is given by 元、=s.m×mg=li×mg(cos0f-sin8)=-lmgsin0k (24.2.1) Following the same steps that led from Equation (24.1.1)to Equation (24.1.4),the rotational equation for the physical pendulum is -mgln sine-Is df d20 (24.2.2) where Is the moment of inertia about the pivot point S.As with the simple pendulum, for small angles sin=,Equation (24.2.2)reduces to the simple harmonic oscillator equation d'e mglem (24.2.3) Is The equation for the angle e(t)is given by 0(t)=Acos(@1)+Bsin(@1), (24.2.4) where the angular frequency is given by 0兰 mg lem (physical pendulum), (24.2.5) and the period is T= 2π (physical pendulum). (24.2.6) 00 mg lem Substitute the parallel axis theorem,I=minto Eq.(24.2.6)with the result that T=2π (physical pendulum). (24.2.7) Thus,if the object is "small"in the sense thatm the expressions for the physical pendulum reduce to those for the simple pendulum.The z-component of the angular velocity is given by 24-3
24-3 The gravitational force acts at the center of mass of the physical pendulum. Denote the distance of the center of mass to the pivot point S by cml . The torque analysis is nearly identical to the simple pendulum. The torque about the pivot point S is given by τ S = rS , cm × m g = l cmrˆ × mg(cosθ rˆ − sinθ ˆ θ) = −l cmmg sinθ kˆ . (24.2.1) Following the same steps that led from Equation (24.1.1) to Equation (24.1.4), the rotational equation for the physical pendulum is − mglcm sinθ = IS d2 θ dt 2 , (24.2.2) where IS the moment of inertia about the pivot point S . As with the simple pendulum, for small angles sinθ ≈θ , Equation (24.2.2) reduces to the simple harmonic oscillator equation d2 θ dt 2 − mglcm IS θ . (24.2.3) The equation for the angle θ(t) is given by θ(t) = Acos(ω0 t)+ Bsin(ω0 t), (24.2.4) where the angular frequency is given by ω0 mg l cm IS (physical pendulum) , (24.2.5) and the period is T = 2π ω0 2π IS mg l cm (physical pendulum). (24.2.6) Substitute the parallel axis theorem, IS = ml cm 2 + Icm , into Eq. (24.2.6) with the result that T 2π l cm g + Icm mg l cm (physical pendulum). (24.2.7) Thus, if the object is “small” in the sense that Icm << ml cm 2 , the expressions for the physical pendulum reduce to those for the simple pendulum. The z-component of the angular velocity is given by
a.(0=B0=-0,4sina,)+,Bcox@,）. (24.2.8) dt The coefficients 4 and B can be determined form the initial conditions by setting t=0 in Egs.(24.2.4)and(24.2.8)resulting in the conditions that A=(1=0)≡0 B=0:1=0=02 (24.2.9) 一三一 00 00 Therefore the nfortheane(r)d))reve by O)=0,cos!)+sin,). (24.2.10) 0.0= ed)=-0,8sin(@,0+0ncos(0,) (24.2.11) 24.3 Worked Examples Example 24.1 Oscillating Rod A physical pendulum consists of a uniform rod of length d and mass m pivoted at one end.The pendulum is initially displaced to one side by a small angle e and released from rest with <<1.Find the period of the pendulum.Determine the period of the pendulum using (a)the torque method and(b)the energy method. pivot P ⊙k *8 Figure 24.3 Oscillating rod (a)Torque Method:with our choice of rotational coordinate system the angular acceleration is given by 24-4
24-4 ω z (t) = dθ dt (t) = −ω0Asin(ω0 t) +ω0Bcos(ω0 t). (24.2.8) The coefficients A and B can be determined form the initial conditions by setting t = 0 in Eqs. (24.2.4) and (24.2.8) resulting in the conditions that A = θ(t = 0) ≡ θ0 B = ω z (t = 0) ω0 ≡ ω z,0 ω0 . (24.2.9) Therefore the equations for the angle θ(t) and ω z (t) = dθ dt (t) are given by θ(t) = θ0 cos(ω0 t) + ω z,0 ω0 sin(ω0 t), (24.2.10) ω z (t) = dθ dt (t) = −ω0 θ0 sin(ω0 t) +ω z,0 cos(ω0 t). (24.2.11) 24.3 Worked Examples Example 24.1 Oscillating Rod A physical pendulum consists of a uniform rod of length d and mass m pivoted at one end. The pendulum is initially displaced to one side by a small angle θ0 and released from rest with θ0 <<1. Find the period of the pendulum. Determine the period of the pendulum using (a) the torque method and (b) the energy method. Figure 24.3 Oscillating rod (a) Torque Method: with our choice of rotational coordinate system the angular acceleration is given by
a= doi. (24.3.1) dr The force diagram on the pendulum is shown in Figure 24.4.In particular,there is an unknown pivot force and the gravitational force acts at the center of mass of the rod. 'Pr 个 m 因E Figure 24.4 Free-body force diagram on rod The torque about the pivot point P is given by 元p=Tpem x mg. (24.3.2) The rod is uniform,therefore the center of mass is a distance d/2 from the pivot point. The gravitational force acts at the center of mass,so the torque about the pivot point P is given by 元p=(d/2)f×mg(-sin66+cosf)=-(d/2)mg sin0k (24.3.3) The rotational equation of motion about P is then =Ia. (24.3.4) Substituting Egs.(24.3.3)and(24.3.1)into Eq.(24.3.4)yields -(d12)mg sin (24.3.5) When the angle of oscillation is small,we may use the small angle approximation sin0≡8，then Eq.(24.3.S)becomes 24-5
24-5 α = d2 θ dt 2 kˆ . (24.3.1) The force diagram on the pendulum is shown in Figure 24.4. In particular, there is an unknown pivot force and the gravitational force acts at the center of mass of the rod. Figure 24.4 Free-body force diagram on rod The torque about the pivot point P is given by τ P = rP,cm × m g . (24.3.2) The rod is uniform, therefore the center of mass is a distance d / 2 from the pivot point. The gravitational force acts at the center of mass, so the torque about the pivot point P is given by τ P = (d / 2)rˆ × mg(−sinθ ˆ θ + cosrˆ) = −(d / 2)mg sinθ kˆ . (24.3.3) The rotational equation of motion about P is then τ P = IP α . (24.3.4) Substituting Eqs. (24.3.3) and (24.3.1) into Eq. (24.3.4) yields −(d / 2)mg sinθ kˆ = IP d2 θ dt 2 kˆ . (24.3.5) When the angle of oscillation is small, we may use the small angle approximation sinθ ≅ θ , then Eq. (24.3.5) becomes
d9,(d/2)mg0=0, 3+ (24.3.6) Ip which is a simple harmonic oscillator equation.The angular frequency of small oscillations for the pendulum is (d/2)mg 00= (24.3.7) Ip The moment of inertia of a rod about the end point P is I=(1/3)md2 therefore the angular frequency is (d/2)mg (312)8 00 (24.3.8) V(1/3)md2 d with period 2d (24.3.9) 0。 (b)Energy Method:Take the zero point of gravitational potential energy to be the point where the center of mass of the pendulum is at its lowest point (Figure 24.5),that is, 0=0. 个 Cm 个 m9(4-co56)1 Figure 24.5 Energy diagram for rod When the pendulum is at an angle 0 the potential energy is U=mg(1-c0s0). (24.3.10) 2 The kinetic energy of rotation about the pivot point is 24-6
24-6 d2 θ dt 2 + (d / 2)mg IP θ 0 , (24.3.6) which is a simple harmonic oscillator equation. The angular frequency of small oscillations for the pendulum is ω0 (d / 2)mg IP . (24.3.7) The moment of inertia of a rod about the end point P is IP = (1/ 3)md2 therefore the angular frequency is ω0 (d / 2)mg (1/ 3)md2 = (3/ 2)g d (24.3.8) with period T = 2π ω0 2π 2 3 d g . (24.3.9) (b) Energy Method: Take the zero point of gravitational potential energy to be the point where the center of mass of the pendulum is at its lowest point (Figure 24.5), that is, θ = 0 . Figure 24.5 Energy diagram for rod When the pendulum is at an angle θ the potential energy is U = mg d 2 (1− cosθ). (24.3.10) The kinetic energy of rotation about the pivot point is
1 (24.3.11) 2 10.2 The mechanical energy is then E=U+Km0=mg -cos)+,0, (24.3.12) with I=(1/3)md2.There are no non-conservative forces acting(by assumption),so the mechanical energy is constant,and therefore the time derivative of energy is zero, 0=de d E=mgsine ae do (24.3.13) d dt I:dt Recall that o.=de/dt and a.=do.dt=d2/dt2,so Eq.(24.3.13)becomes 0=0 m8号s血0+1， d (24.3.14) There are two solutions,=0,in which case the rod remains at the bottom of the swing, 0=m82sin8+1。 d28 dr (24.3.15) Using the small angle approximation,we obtain the simple harmonic oscillator equation (Eq.(24.3.6) de mg(d12）0=0. d2 (24.3.16) I Example 24.3 Torsional Oscillator A disk with moment of inertia about the center of mass Im rotates in a horizontal plane. It is suspended by a thin,massless rod.If the disk is rotated away from its equilibrium position by an angle 0,the rod exerts a restoring torque about the center of the disk with magnitude given by=b(Figure 24.6),where b is a positive constant.At t=0,the disk is released from rest at an angular displacement of 0.Find the subsequent time dependence of the angular displacement e(t). 24-7
24-7 Krot = 1 2 I p ω z 2 . (24.3.11) The mechanical energy is then E = U + Krot = mg d 2 (1− cosθ ) + 1 2 I p ω z 2 , (24.3.12) with 2 (1/ 3) PI = md . There are no non-conservative forces acting (by assumption), so the mechanical energy is constant, and therefore the time derivative of energy is zero, 0 = dE dt = mg d 2 sinθ dθ dt + I p ω z dω z dt . (24.3.13) Recall that ω z = dθ / dt and α z = dω z / dt = d2 θ / dt 2 , so Eq. (24.3.13) becomes 0 = ω z mg d 2 sinθ + I p d2 θ dt 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ . (24.3.14) There are two solutions, ω z = 0 , in which case the rod remains at the bottom of the swing, 0 = mg d 2 sinθ + I p d2 θ dt 2 . (24.3.15) Using the small angle approximation, we obtain the simple harmonic oscillator equation (Eq. (24.3.6)) d2 θ dt 2 + mg(d / 2) I p θ 0 . (24.3.16) Example 24.3 Torsional Oscillator A disk with moment of inertia about the center of mass Icm rotates in a horizontal plane. It is suspended by a thin, massless rod. If the disk is rotated away from its equilibrium position by an angle θ , the rod exerts a restoring torque about the center of the disk with magnitude given by τ cm = bθ (Figure 24.6), where b is a positive constant. At t = 0 , the disk is released from rest at an angular displacement of θ0 . Find the subsequent time dependence of the angular displacement θ (t)
Figure 24.6 Example 24.3 with exaggerated angle 0 Solution:Choose a coordinate system such that k is pointing upwards(Figure 24.6), then the angular acceleration is given by d d= k (24.3.17) dr The torque about the center of mass is given in the statement of the problem as a restoring torque,therefore 元m=-b0k (24.3.18) The z-component of the rotational equation of motion is -b0-Icm di (24.3.19) This is a simple harmonic oscillator equation with solution e(1)=Acos(@1)+Bsin(@t) (24.3.20) where the angular frequency of oscillation is given by 0。=Vb11m 24.3.21) The z-component of the angular velocity is given by @.)=d ()=-0Asin(0,)+0Bcos(0). (24.3.22) dt The initial conditions at t=0,are (t=0)=A=0,and (de/dr)(t=0)=0B=0. Therefore θ()=0。cos(Vb1Lmt). (24.3.23) 24-8
24-8 Figure 24.6 Example 24.3 with exaggerated angle θ Solution: Choose a coordinate system such that kˆ is pointing upwards (Figure 24.6), then the angular acceleration is given by α = d2 θ dt 2 kˆ . (24.3.17) The torque about the center of mass is given in the statement of the problem as a restoring torque, therefore τcm = −bθ kˆ . (24.3.18) The z-component of the rotational equation of motion is −bθ = Icm d2 θ dt 2 . (24.3.19) This is a simple harmonic oscillator equation with solution θ(t) = Acos(ω0 t) + Bsin(ω0 t) (24.3.20) where the angular frequency of oscillation is given by ω0 = b / Icm . (24.3.21) The z -component of the angular velocity is given by ω z (t) = dθ dt (t) = −ω0 Asin(ω0 t) +ω0Bcos(ω0 t). (24.3.22) The initial conditions at t = 0 , are 0 θ (t = 0) = A = θ , and (dθ / dt)(t = 0) = ω0B = 0 . Therefore θ(t) = θ0 cos( b / Icm t). (24.3.23)
Example 24.4 Compound Physical Pendulum A compound physical pendulum consists of a disk of radius R and mass m,fixed at the end of a rod of mass m and length /(Figure 24.7a).(a)Find the period of the pendulum. (b)How does the period change if the disk is mounted to the rod by a frictionless bearing so that it is perfectly free to spin? mg (a) (b) Figure 24.7(a)Example 24.4 (b)Free-body force diagram Solution:We begin by choosing coordinates.Let k be normal to the plane of the motion of the pendulum pointing out of the plane of the Figure 24.7b.Choose an angle variable 0 such that counterclockwise rotation corresponds to a positive z-component of the angular velocity.Thus a torque that points into the page has a negative z-component and a torque that points out of the page has a positive z-component.The free-body force diagram on the pendulum is also shown in Figure 24.7b.In particular,there is an unknown pivot force,the gravitational force acting at the center of mass of the rod,and the gravitational force acting at the center of mass of the disk.The torque about the pivot point is given by 元p=Tpem X m,g+Tpdisk X ma8 (24.3.24) Recall that the vectorm points from the pivot point to the center of mass of the rod a distance 1/2 from the pivot.The vector points from the pivot point to the center of mass of the disk a distance from the pivot.Torque diagrams for the gravitational force on the rod and the disk are shown in Figure 24.8.Both torques about the pivot are in the negative k-direction (into the plane of Figure 24.8)and hence have negative = components, =-(m(1/2)+m l)gsinek (24.3.25) 24-9
24-9 Example 24.4 Compound Physical Pendulum A compound physical pendulum consists of a disk of radius R and mass md fixed at the end of a rod of mass mr and length l (Figure 24.7a). (a) Find the period of the pendulum. (b) How does the period change if the disk is mounted to the rod by a frictionless bearing so that it is perfectly free to spin? (a) (b) Figure 24.7 (a) Example 24.4 (b) Free-body force diagram Solution: We begin by choosing coordinates. Let kˆ be normal to the plane of the motion of the pendulum pointing out of the plane of the Figure 24.7b. Choose an angle variable θ such that counterclockwise rotation corresponds to a positive z-component of the angular velocity. Thus a torque that points into the page has a negative z-component and a torque that points out of the page has a positive z-component. The free-body force diagram on the pendulum is also shown in Figure 24.7b. In particular, there is an unknown pivot force, the gravitational force acting at the center of mass of the rod, and the gravitational force acting at the center of mass of the disk. The torque about the pivot point is given by τ P = rP,cm × mr g + rP,disk × md g . (24.3.24) Recall that the vector rP,cm points from the pivot point to the center of mass of the rod a distance l / 2 from the pivot. The vector rP,disk points from the pivot point to the center of mass of the disk a distance l from the pivot. Torque diagrams for the gravitational force on the rod and the disk are shown in Figure 24.8. Both torques about the pivot are in the negative kˆ -direction (into the plane of Figure 24.8) and hence have negative zcomponents, τ P = −(mr (l / 2) + md l)g sinθ kˆ . (24.3.25)